Solving third order nonlinear boundary value problem

6 Ansichten (letzte 30 Tage)
parham kianian
parham kianian am 21 Dez. 2020
Bearbeitet: Torsten am 1 Okt. 2023
I have a boundary value problem in this form:
where alpha is a nonzero constant and boudary conditions are:
f(-1) = 0
f(0) = 1
f(1) = 0
How can I solve this problem using bvp5c?
  5 Kommentare
PRITESH
PRITESH am 1 Okt. 2023
Hi @Torsten. Thanks for the reply. I am still confused while applying this technique. I am trying to use bvp4c and am getting singular Jacobian values.
Torsten
Torsten am 1 Okt. 2023
Bearbeitet: Torsten am 1 Okt. 2023
I am trying to use bvp4c and am getting singular Jacobian values.
I don't:
alpha = 1;
xc = 0;
xmesh = [linspace(-1,xc,1000),linspace(xc,1,1000)];
solinit = bvpinit(xmesh, [0 0 0]);
sol = bvp5c(@(x,y,r)f(x,y,r,alpha),@bc,solinit);
plot(sol.x,sol.y(1,:))
function dydx = f(x,y,region,alpha)
dydx = [y(2);y(3);-alpha*y(1)*y(2)-4*alpha^2*y(2)];
end
function res = bc(yl,yr)
res = [yl(1,1);yr(1,1)-1;yr(1,1)-yl(1,2);yr(2,1)-yl(2,2);yr(3,1)-yl(3,2);yr(1,2)];
end

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Kategorien

Mehr zu Robotics System Toolbox finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by